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There are $n$ bins of which the $k^{th}$ contains $k −1$ blue balls and $n−k$ red balls. You pick a bin at random and remove two balls at random without replacement. Find the probability that:

(a) the second ball is red

(b) the second ball is red, given that the first is red.

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For (a), the answer is $\frac12$ by symmetry.

For (b), the conditional probability that we're drawing from the $k$-th bin is proportional to the number $n-k$ of red balls, and the probability of drawing another red ball is the number $n-k-1$ of red balls left over the total number $n-2$ of balls left, yielding

$$ \sum_k\frac{n-k}{\sum_k(n-k)}\frac{n-k-1}{n-2}=\frac1{\sum_k(n-k)}\frac1{n-2}\cdot2\sum_k\binom{n-k}2=\frac{\binom n3}{\binom n2}\frac2{n-2}=\frac23. $$

I suspect there's a nicer way to show that.

P.S.: Here's the nicer way. Instead of drawing from $n$ bins with $n-1$ precoloured balls, imagine that you have one bin with $n$ white balls. Arrange the balls linearly and randomly uniformly select one of them to serve as the boundary between red and blue. Colour the balls before it red and the ones after it blue. This is equivalent to selecting one of the $n$ bins with $n-1$ precoloured balls. Now select two more balls. All that matters for the outcome is the relative order of the three selected balls. There are $6$ possible orders, and they're all equally likely, so the solution to (b) is now a simple matter of counting in how many of them the first ball is red ($3$) and in how many of those the second ball is also red ($2$).

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  • $\begingroup$ +1 for the nicer way. $\endgroup$ – saulspatz Sep 1 '18 at 14:12
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You pick a bin at random, so each bin has equal probability $(\frac 1n)$ of being picked, and each bin contains $n-1$ balls.

(a) We want the probability that we pick a red ball, which for bin $i$ is $\frac{n-i}{n-1}$, so the total $$p_a = \frac 1n \sum_{i=1}^n \frac{n-i}{n-1} = \frac{1}{n(n-1)} \left( n^2 - \frac{n(n+1)}{2} \right) = \frac{1}{n(n-1)}\frac{n^2-n}{2} = \frac 12.$$

(b) As above, except you want both balls to be red. So for bin $i$ you have $\frac{n-i}{n-1}$ for the first ball, and $\frac{n-i-1}{n-2}$ for the second ball. The probability is then $$ \frac{\sum_{i=1}^n (n-i)(n-i-1)}{n(n-1)(n-2)} = \frac{\sum_{j=0}^{n-1} j^2 - j}{n(n-1)(n-2)} = \frac{ \frac{(n-1)n(2n-1)}{6} - \frac{(n-1)(n)}{2}}{n(n-1)(n-2)} = \frac{ \frac{(2n-1)}{6} - \frac{1}{2}}{(n-2)} = \frac 13.$$

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  • $\begingroup$ This is the probability for picking two red balls. The question was about the probability that the second ball is red, given that the first is red. $\endgroup$ – joriki Aug 13 '15 at 14:17
  • $\begingroup$ My bad: I misunderstood the question. $\endgroup$ – Maciek Aug 13 '15 at 14:51

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